What's Possible?

742 /www/nrich/html/content/00/11/six5/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> Many numbers can be expressed as the difference of two perfect squares. For example, $$20 = 6^2 - 4^2$$ $$21 = 5^2 - 2^2$$ How many of the numbers from $1$ to $20$ can you express as the difference of two perfect squares? <h3>Here are some questions to consider:</h3> What do you notice about the difference between squares of consecutive numbers? What about the difference when I square two numbers which differ by $2$? By $3$? By $4$...? When is the difference between two square numbers odd? And when is it even? What do you notice about the numbers you CANNOT make? <div>Can you prove any of your findings?</div> You may want to take a look at <a href="http://nrich.maths.org/658&part=" style="font-style: italic;">Plus Minus</a> next. </mdoxml> <mdoxml version="1.0"> Hannah from Munich International School worked out which numbers between 1 and 20 were possible: 1: $1^2 - 0^2$ is $1 - 0 = 1$ 2: does not work 3: $2^2 - 1^2$ is $4 - 1 = 3$ 4: $2^2 - 0^2$ is $4 - 0 = 4$ 5: $3^2 - 2^2$ is $9 - 4 = 5$ 6: does not work 7: $4^2 - 3^2$ is $16 - 9 = 7$ 8: $3^2 - 1^2$ is $9 - 1 = 8$ 9: $5^2 - 4^2$ is $25 - 16 = 9$ 10: does not work 11: $6^2 - 5^2$ is $36 - 25 = 11$ 12: $4^2 - 2^2$ is $16 - 4 = 12$ 13: $7^2 - 6^2$ is $49 - 36 = 13$ 14: does not work 15: $8^2 - 7^2$ is $64 - 49 = 15$ 16: $5^2 - 3^2$ is $25 - 9 = 16$ 17: $9^2 - 8^2$ is $81 - 64 = 17$ 18: does not work 19: $10^2 - 9^2$ is $100 - 81 = 19$ 20: $6^2 - 4^2$ is $36 - 16 = 20$ You can express 16 of the 20 numbers as a difference of two perfect squares. A pattern occuring throughout these solutions is that all odd numbers can be represented by a difference of two perfect squares, as well as all numbers resulting in an integer when divided by four. Joshua from St John's School used algebra to show how odd numbers and multiples of four could be made: You can make every odd number by taking consecutive squares. $(n+1)^2 - n^2 = 2n+1$, every odd number can be written in the form $2n+1$. Similarly, you can make every multiple of 4 by taking squares with a difference of 2. All other numbers you can't make: $(n+x)^2 - n^2 = x^2 + 2nx = x(x+2n)$ If $x$ is odd then $x^2$ is also odd and $2nx$ is even. An odd plus an even is odd. If $x$ is even then $x^2$ is a multiple of $4$ and $2nx$ is also a multiple of $4$ so $(n+x)^2 - n^2$ is a multiple of $4$. Numbers such as 2,6,10 etc can't be made because these are made by multiplying an even by an odd. Patrick from Woodbridge School and Richard from Mearns Castle School both used a formula for the difference of two squares to investigate which numbers were possible. Here is Patrick's solution: $a^2 - b^2= (a + b)(a - b)$ Thus, a number $n$ can only be a difference of two squares if it has two factors of the form $(a + b)$ and $(a - b)$, where $a + b \geq \sqrt{n}$ and $a - b \leq \sqrt{n}$. If $1$ could be written as the difference of squares then $(a + b) = (a - b) = 1$, as the only factor of $1$ is $1$. Thus, $a = 1$ and $b = 0$. So $1 = 1^2 - 0^2$. If $2$ could be written as the difference of two squares then $(a + b) = 2$ and $(a - b) = 1$, as $2$ is prime, and clearly $a - b < a + b$ for positive $b$. Thus, solving the simultaneous equations, $b = 2 - a$, so $a - (2 - a) = 1$, $2a = 3$ and $a = \frac{3}{2}$. Thus, $2$ is not representable as the difference of two integer squares. This method can be generalised for any prime $p$ by solving $(a + b) = p$ and $(a - b) = 1$ Adding the equations gives $2a = p + 1$, so $a = \frac{p+1}{2}$. Since $a - b = 1$, $b = a-1$, so $b = \frac{p+1}{2} - 1 = \frac{p-1}{2}$. Thus, any odd prime can be written as the difference of two squares. Any square number $n$ can also be written as the difference of two squares, by taking $a = \sqrt{n}$ and $b = 0$. Generally, a number can be written as the difference of two squares if it has two factors of the same parity, since if $a + b$ is odd and $a - b$ is even, when the two equations are added we would get $2a$ odd, so the solution would not be an integer. So a number cannot be written as a difference of two squares if and only if it is equivalent to $2$ mod $4$ (leaves a remainder of $2$ when divided by $4$). Richard from Comberton Village College considered prime factorisation: An integer $x$ can be written as the difference between two square integers unless it contains only one $2$ in its prime power factorisation. Proof: $x=a^2 - b^2$ can be rewritten in the form $(a+b)(a-b)$. Suppose $u$ and $v$ are two factors of $x$ such that $u = a + b$ and $v = a - b$. Subtracting the two equations, we get $u - v = 2b$. This tells us that the difference between the two factors will always be even if the number can be written in this form. Now, to get an even difference, we either need $u$ and $v$ to be both odd, or both even. Therefore, any number with either two even or odd factors can be written in this form. This leaves numbers which will always have one odd and one even factor. If $v$ is an odd factor of $x$, it must have a prime factorisation of only odd numbers. Suppose $u$ is an even factor of $x$ and contains more than one $2$ in its prime factorisation. This means that $u$ could be divided by two and still be even, and at the same time $v$ could be doubled so that it too is even. This means that $x$ has a factor pair which are both even, and so can be written as the difference of two squares. If both factors are odd an even difference will also be attained, so if there are no twos in the prime factorisation of $x$, then x can be written as the difference of two squares. This leaves the case where there is exactly one $2$ in the prime factorisation. This means that one factor must contain the $2$ and not the other, and that the factors cannot be manipulated in the fashion described before. This will always leave factors with an odd difference suggesting $x$ cannot be written as the difference of two squares. This also gives a very nice algorithm for finding two squares which have a difference of $x$: 1. Establish how many $2$s are in the prime factorisation of $x$. If there is only one, x cannot be written as the difference of two squares. 2. Choose a factor pair which are both even or both odd, call them $u$ and $v$. Find their difference and divide it by 2 to find the smaller square. $\frac{u - v}{2}=b$ from before. 3. Square $b$ and add it to $x$. Square root the sum and that is $a$. We now have $x = a^2 - b^2$. For example, $60 = 2 \times 2 \times 3 \times 5$. There is more than one 2 so it is possible. The two factors $2 \times 5 = 10$ and $2 \times 3 = 6$ give a difference of 4. Halve it to get $b = 2$. $x + b^2 = 60 + 4 = 64$. $\sqrt{64} = 8$, so $a = 8$. This gives us $60 = 8^2 - 2^2$. We could also have taken $u = 2 \times 3 \times 5$ and $v = 2$, which gives a difference of 28. Halve it to get $b=14$. $x + b^2 = 60 + 196 = 256$. $\sqrt{256} = 16$ so $a = 16$. $16^2 - 14^2 = 60$ </mdoxml> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/742">This problem</a> starts by asking students to find which numbers can be expressed as the difference of two square numbers, and then suggests some possible avenues for exploration. This can then be used as a springboard to generalisations and the use of algebra for justifications and proof. Along the way, students have the opportunity to make use of the important identity $a^2 - b^2 = (a + b)(a - b)$. <h3>Possible approach</h3> <div>Arrange the class in groups of two or three, and hand each group one set from <a href="/content/00/11/six5/What%27s%20Possible.doc">these cards</a>. Ask them to work in their groups for the first few minutes, looking at what they notice about their three cards, whether they could create other cards that would belong in the same set, and what questions are prompted by them.</div> <div> </div> <div>Next, bring the class together to share what was on their cards, and what questions occurred to them. Do any of their questions resolve themselves once they see someone else's cards? Make sure conjectures and questions are noted on the board.</div> <div> </div> <div>One way to begin to resolve these conjectures and provoke new ones is to challenge students to find numbers between $1$ and $30$ which can be expressed as the difference between two square numbers. Encourage them to find more than one solution where possible, and draw attention to systematic ways of working and recording. Once everyone has had some time to find most of the answers, ask students to share what they have found and tabulate the answers on the board. Note down any new conjectures that emerge at this point.</div> <div> </div> <div>Allow students to choose a line of enquiry, working on their own or in small groups. The emphasis should be on proving any conjectures they make, whether using diagrams or more formal algebraic methods. A diagrammatic method for calculating the difference of two squares is explored in the problem <a href="http://nrich.maths.org/658&part=">Plus Minus</a>.</div> The plenary can involve students presenting their findings to the rest of the class. Expect students to be clear and rigorous in their justifications. Encourage students to challenge any proofs that lack clarity and rigour, and suggest ways of improving them. One way to round off the work on this problem could be: "In a while, I'm going to give you a number and ask you to quickly find one or more ways to write it as the difference of two squares, or to convince me that it can't be done." <h3>Key questions</h3> <div>What do you notice about the difference between squares of consecutive numbers?</div> <div>What about the difference when I square two numbers which differ by $2$? By $3$? By $4$ ...?</div> <div>When is the difference between two square numbers odd?</div> <div>And when is it even?</div> <div>What do you notice about the numbers you CANNOT make?</div> <div>Can you prove any of your findings?</div> <h3>Possible extension</h3> <div>Every odd prime number can be written as the difference of two squares. Prove that there is only one way to write an odd prime number as the difference of two squares, and that this is the difference of two successive squares.</div> <div>120 can be written as the difference of the squares of whole numbers in the following ways:</div> <div>$$31^2-29^2$$</div> <div>$$17^2-13^2$$</div> <div>$$13^2-7^2$$</div> <div>$$11^2-1^2$$</div> <div>Can you find all the possible ways to write 924 as the difference of the squares of two whole numbers? Can you prove that you have found them all?</div> <div>*REALLY CHALLENGING*</div> <div>Can you describe a way for working out how many ways there are to write any number as the difference of two squares?</div> <h3>Possible support</h3> <div>Encourage students to list their results in a systematic way, gathering together in separate columns numbers made from squaring two numbers which differ by $1$, $2$, $3$... as started below:</div> <table border="1"> <tbody> <tr> <td> </td> <td style="text-align: center;">1</td> <td style="text-align: center;">2</td> <td style="text-align: center;">3</td> </tr> <tr> <td>1</td> <td>$1^2-0^2$</td> <td> </td> <td> </td> </tr> <tr> <td>2</td> <td> </td> <td> </td> <td> </td> </tr> <tr> <td>3</td> <td>$2^2-1^2$</td> <td> </td> <td> </td> </tr> <tr> <td>4</td> <td> </td> <td>$2^2-0^2$</td> <td> </td> </tr> <tr> <td>5</td> <td>$3^2-2^2$</td> <td> </td> <td> </td> </tr> <tr> <td>6</td> <td> </td> <td> </td> <td> </td> </tr> <tr> <td>7</td> <td>$4^2-3^2$</td> <td> </td> <td> </td> </tr> <tr> <td>8</td> <td> </td> <td>$3^2-1^2$</td> <td> </td> </tr> <tr> <td>9</td> <td>$5^2-4^2$</td> <td> </td> <td>$3^2-0^2$</td> </tr> </tbody> </table> </mdoxml> <mdoxml version="1.0">You may find this image helpful when proving your findings. <mdo:image height="280" width="282" src="whatsposs.png" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="ISO-8859-1"?> <mdoxml version="1.0"> We have to find all possible integer values of a and b where the difference of the squares of $a$ and $b$ is $924$ and then we have to discover which numbers CANNOT be written as the difference of two perfect squares? The use of a spreadsheet lends itself to systematic exploration of number, in this case squares and differences of squares. Students from Bourne Grammar School who worked hard on this question would have gone further had they used spreadsheets. We know that $$(a^2 - b^2) = (a - b)(a + b)$$ It is always true that $(a - b)$ and $(a + b)$ must both be even or both odd. If they are both even then the product is a multiple of $4$. If they are both odd then the product is odd. So even numbers which are not multiples of $4$ cannot be written as the difference of two squares. As $924 = 2 \times2 \times3 \times7 \times11$ we know that $(a - b)$ and $(a + b)$ must both be even. From this we can write down and solve pairs of simple equations to find $a$ and $b$ using all the combinations of prime factors of $924$ which give pairs of even factors. The solution below from Soh Yong Sheng, Raffles Institution, Singapore uses a different method based on the fact that the square of any number is the sum of consecutive odd numbers $$n^2 = 1 + 3 + 5 + ... (2n - 1)$$ and so the difference of two squares is the sum of consecutive odd numbers. Many numbers can be expressed as differences of 2 squares, but exactly which? It is known that $1=1 (1^2)$ $1+3=4 (2^2)$ $1+3+5=9 (3^2)$ $1+3+5+7=16 (4^2)$ $\ldots$ $1+3+\ldots+n= ((n+1)/2)^2 $ Using this, we try to find the sums of consecutive odd numbers that add up to $924$ which will tell us $a$ and $b$. Case of $2$ odds $924 = 461 + 463$ $a= 464/2 = 232$, $a^2 = 53824$, $b= 230$, $b^2 = 52900$. One possible solution. Case of $4$ odds $924 = 229 + 231 + 231 + 233$ so this is impossible. Case of $6$ odds $924=149+151+153+155+157+159$ $a = (159 + 1)/2 = 80$, $a^2 = 6400$, $b = (149-1)/2 = 74$, $b^2 = 5476$. A possible solution. Cases of $8$, $10$ and $12$ odds impossible Case of $14$ odds $924=53+55+57+59+61+63+65+67+69+71+73+75+77+79$ $a= 80/2 = 40$, $a^2 = 1600$, $b = (53-1)/2= 26$, $262 = 676$. Another solution. Cases of $16$, $18$ and $20$ odds are impossible Case of $22$ odds $924=21+23+25+27+29+31+33+35+37+39+41+43+45+47+49+51+53+55+57+59+61+63$ $a = 64/2 =32$, $a^2 = 1024$, $b = (21-1)/2$, $b^2 = 100$ Another solution. This gives the pairs $(232, 230)$, $(80, 74)$, $(26, 40)$ and $(64,10)$ The solutions can also be negative integers $a = \pm232$ and $b = \pm 230$ etc. giving sixteen solutions in all. </mdoxml> 2 4 0 0 0 1 0 What's Possible? Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make? Algebra Creating expressions/formulae Algebra Using symbols Using, Applying and Reasoning about Mathematics Generalising Algebra Difference of two squares Using, Applying and Reasoning about Mathematics Making and proving conjectures Information and Communications Technology smartphone